This is just the following result.

Lemma 33.23.1. Let $k$ be an algebraically closed field. Let $X$ be a proper $k$-scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $V \subset H^0(X, \mathcal{L})$ be a $k$-subvector space. If

for every pair of distinct closed points $x, y \in X$ there is a section $s \in V$ which vanishes at $x$ but not at $y$, and

for every closed point $x \in X$ and nonzero tangent vector $\theta \in T_{X/k, x}$ there exists a section $s \in V$ which vanishes at $x$ but whose pullback by $\theta $ is nonzero,

then $\mathcal{L}$ is very ample and the canonical morphism $\varphi _{\mathcal{L}, V} : X \to \mathbf{P}(V)$ is a closed immersion.

**Proof.**
Condition (1) implies in particular that the elements of $V$ generate $\mathcal{L}$ over $X$. Hence we get a canonical morphism

\[ \varphi = \varphi _{\mathcal{L}, V} : X \longrightarrow \mathbf{P}(V) \]

by Constructions, Example 27.21.2. The morphism $\varphi $ is proper by Morphisms, Lemma 29.41.7. By (1) the map $\varphi $ is injective on closed points (computation omitted). In particular, the fibre over any closed point of $\mathbf{P}(V)$ is a singleton (small detail omitted). Thus we see that $\varphi $ is finite, for example use Cohomology of Schemes, Lemma 30.21.2. To finish the proof it suffices to show that the map

\[ \varphi ^\sharp : \mathcal{O}_{\mathbf{P}(V)} \longrightarrow \varphi _*\mathcal{O}_ X \]

is surjective. This we may check on stalks at closed points. Let $x \in X$ be a closed point with image the closed point $p = \varphi (x) \in \mathbf{P}(V)$. Since $\varphi ^{-1}(\{ p\} ) = \{ x\} $ by (1) and since $\varphi $ is proper (hence closed), we see that $\varphi ^{-1}(U)$ runs through a fundamental system of open neighbourhoods of $x$ as $U$ runs through a fundamental system of open neighbourhoods of $p$. We conclude that on stalks at $p$ we obtain the map

\[ \varphi ^\sharp _ x : \mathcal{O}_{\mathbf{P}(V), p} \longrightarrow \mathcal{O}_{X, x} \]

In particular, $\mathcal{O}_{X, x}$ is a finite $\mathcal{O}_{\mathbf{P}(V), p}$-module. Moreover, the residue fields of $x$ and $p$ are equal to $k$ (as $k$ is algebraically closed – use the Hilbert Nullstellensatz). Finally, condition (2) implies that the map

\[ T_{X/k, x} \longrightarrow T_{\mathbf{P}(V)/k, p} \]

is injective since any nonzero $\theta $ in the kernel of this map couldn't possibly satisfy the conclusion of (2). In terms of the map of local rings above this means that

\[ \mathfrak m_ p/\mathfrak m_ p^2 \longrightarrow \mathfrak m_ x/\mathfrak m_ x^2 \]

is surjective, see Lemma 33.16.5. Now the proof is finished by applying Algebra, Lemma 10.20.3.
$\square$

Lemma 33.23.2. Let $k$ be an algebraically closed field. Let $X$ be a proper $k$-scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Suppose that for every closed subscheme $Z \subset X$ of dimension $0$ and degree $2$ over $k$ the map

\[ H^0(X, \mathcal{L}) \longrightarrow H^0(Z, \mathcal{L}|_ Z) \]

is surjective. Then $\mathcal{L}$ is very ample on $X$ over $k$.

**Proof.**
This is a reformulation of Lemma 33.23.1. Namely, given distinct closed points $x, y \in X$ taking $Z = x \cup y$ (viewed as closed subscheme) we get condition (1) of the lemma. And given a nonzero tangent vector $\theta \in T_{X/k, x}$ the morphism $\theta : \mathop{\mathrm{Spec}}(k[\epsilon ]) \to X$ is a closed immersion. Setting $Z = \mathop{\mathrm{Im}}(\theta )$ we obtain condition (2) of the lemma.
$\square$

## Comments (2)

Comment #7149 by Manolis C. Tsakiris on

Comment #7299 by Johan on